/*
在一个 8 x 8 的棋盘上，有一个白色车（rook）。也可能有空方块，白色的象（bishop）和黑色的卒（pawn）。它们分别以字符 “R”，“.”，“B” 和 “p” 给出。大写字符表示白棋，小写字符表示黑棋。

车按国际象棋中的规则移动：它选择四个基本方向中的一个（北，东，西和南），然后朝那个方向移动，直到它选择停止、到达棋盘的边缘或移动到同一方格来捕获该方格上颜色相反的卒。另外，车不能与其他友方（白色）象进入同一个方格。

返回车能够在一次移动中捕获到的卒的数量。
 

示例 1：



输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释：
在本例中，车能够捕获所有的卒。
示例 2：



输入：[[".",".",".",".",".",".",".","."],[".","p","p","p","p","p",".","."],[".","p","p","B","p","p",".","."],[".","p","B","R","B","p",".","."],[".","p","p","B","p","p",".","."],[".","p","p","p","p","p",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：0
解释：
象阻止了车捕获任何卒。
示例 3：



输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释： 
车可以捕获位置 b5，d6 和 f5 的卒。
 

提示：

board.length == board[i].length == 8
board[i][j] 可以是 'R'，'.'，'B' 或 'p'
只有一个格子上存在 board[i][j] == 'R'

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/available-captures-for-rook
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
【春招】
*/

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int nRetCount = 0;
        int nWidth = board.size();
        int nHeight = board[0].size();

        int nRockI = 0, nRockJ = 0;
        for (int i = 0, j = 0; i < nWidth; ++i)
        {
            for (j = 0; j < nHeight; ++j)
            {
                if (board[i][j] == 'R')
                {
                    nRockI = i;
                    nRockJ = j;
                    break;
                }
            }
            if (j < nHeight)
            {
                break;
            }
        }

        int nTempRockI, nTempRockJ;

        nTempRockI = nRockI;
        nTempRockJ = nRockJ;
        do { nTempRockI--; } while(nTempRockI >= 0 && board[nTempRockI][nRockJ] == '.');
        if (nTempRockI >= 0 && board[nTempRockI][nRockJ] == 'p')
        {
            nRetCount++;
        }

        nTempRockI = nRockI;
        nTempRockJ = nRockJ;
        do { nTempRockI++; } while(nTempRockI <= 7 && board[nTempRockI][nRockJ] == '.');
        if (nTempRockI <= 7 && board[nTempRockI][nRockJ] == 'p')
        {
            nRetCount++;
        }

        nTempRockI = nRockI;
        nTempRockJ = nRockJ;
        do { nTempRockJ--; } while(nTempRockJ >= 0 && board[nRockI][nTempRockJ] == '.');
        if (nTempRockJ >= 0 && board[nRockI][nTempRockJ] == 'p')
        {
            nRetCount++;
        }

        nTempRockI = nRockI;
        nTempRockJ = nRockJ;
        do { nTempRockJ++; } while(nTempRockJ <= 7 && board[nRockI][nTempRockJ] == '.');
        if (nTempRockJ <= 7 && board[nRockI][nTempRockJ] == 'p')
        {
            nRetCount++;
        }

        return nRetCount;
    }
};